# limits indeterminate form

• November 8th 2007, 04:15 PM
cowboys111
limits indeterminate form
$
\displaystyle\lim_{x\to\infty}
$

(1-1/x)^5x

thank you
• November 8th 2007, 04:50 PM
Soroban
Hello, cowboys111!

I believe you are expected to know that: . $\lim_{u\to\infty}\left(1 + \frac{a}{u}\right)^u \;=\;e^a$

Quote:

$\displaystyle\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^{5x}$

We have: . $\lim_{x\to\infty}\left[\left(1 + \frac{\text{-}1}{x}\right)^x\right]^5 \;=\;\left[\lim_{x\to\infty}\left(1 + \frac{\text{-}1}{x}\right)^x\right]^5\;=\;\left[e^{-x}\right]^5 \;=\;e^{-5x}
$

• November 8th 2007, 05:12 PM
Plato
I sure that it is a typo.
The limit is $e^{-5}$.
• November 8th 2007, 06:13 PM
cowboys111
ok thats easy enough i just didnt know that http://www.mathhelpforum.com/math-he...965b6bd3-1.gif
thanks for the help